Option A : $$\mathrm{Na}_{3}\left[\mathrm{CoCl}_{6}\right]$$

This is an octahedral complex, but it is not diamagnetic. Co in this complex is in +3 oxidation state, and the d-electron configuration is $t_{2g}^6 e_{g}^0$, leading to two unpaired electrons.

Option B : $$\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$$

This is also an octahedral complex, but it is not diamagnetic. The Co ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^5 e_{g}^2$, leading to 3 unpaired electrons.

Option C : $$\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$$

This complex is both octahedral and diamagnetic. Co in this complex is in +3 oxidation state and the d-electron configuration is $t_{2g}^6 e_{g}^0$, which means no unpaired electrons. This complex is more stable due to a strong field ligand (CN^-) which leads to a greater splitting of d-orbitals and more pairing of electrons.

Option D : $$\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}$$

This is an octahedral complex, and it is diamagnetic. The Ni ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^8 e_{g}^0$, leading to no unpaired electrons. However, NH₃ is a weaker field ligand compared to CN^-, so the complex is less stable than Option C.

Hence, the complex that is octahedral, diamagnetic, and the most stable among the given options is $$\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$$.